Line loss itself is a comparison of two powers, those entering and
leaving the line. The line loss in watts is the power entering the line
at the input end minus the power leaving the line at the output end.
Line loss in dB is 10 times the base 10 logarithm of the power entering
the line at the input end divided by the power leaving the line at the
output end. These definitions can be and are used for any two-port device.

If you truly didn't know this, it's surely because you've befuddled
yourself with your forward and reflected waves of average power to the
point that you've lost complete track of the fundamentals.

Roy Lewallen, W7EL

Cecil Moore wrote:

Richard Clark wrote:

You are missing coming to terms with Bart's lesson - still.


It's a pretty simple question that, so far, no one has
answered, not even Bart. dB is always a comparison of
one thing to another. To what is the line loss in watts
being compared?
--
73, Cecil http://www.qsl.net/w5dxp

Roy Lewallen wrote:
Line loss itself is a comparison of two powers, those entering and
leaving the line. The line loss in watts is the power entering the line
at the input end minus the power leaving the line at the output end.

Here's an example: The source is a signal generator equipped with a
circulator-resistor that dissipates all reflected power.

100w SGCR--------------feedline------------------mismatched load

The signal generator is sourcing 100 watts. The load is dissipating
25 watts. The circulator resistor is dissipating 50 watts. The
feedline is dissipating 25 watts. What is the feedline loss in dB?

Is that 25 watts lost from the signal generator output power of
100 watts or lost from the NET power available which is 50 watts?
--
73, Cecil http://www.qsl.net/w5dxp

On Wed, 01 Dec 2004 16:50:00 -0600, Cecil Moore
wrote:
What is the feedline loss in dB?
The same. You haven't substituted it with a lightbulb have you?

Richard Clark wrote:

Cecil Moore wrote:
What is the feedline loss in dB?

The same. You haven't substituted it with a lightbulb have you?

The same as what? The feedline loss is 25 watts. What is the feedline
loss in dB? Please give me a number instead of a hard time.

A signal generator is pouring 100 watts into the line because its
source is seeing 50 ohms. The circulator load resistor is dissipating
50 watts of that source power. The load is dissipating 25 watts and
the feedline is dissipating 25 watts. What is the feedline loss in dB?

Is the dB loss in the feedline calculated using the signal generator
output power of 100 watts or is it calculated using the 50 watts of
NET power available as it would be (by definition) if the source were
a ham transmitter?

Why is it so hard to get a straight answer to this simple question?
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
. . .
Why is it so hard to get a straight answer to this simple question?

The answer is simple. 25 watts.

Roy Lewallen, W7EL

Roy Lewallen wrote:
Cecil Moore wrote:
Why is it so hard to get a straight answer to this simple question?

The answer is simple. 25 watts.

That was a given, Roy. The question is: what is the feedline
loss in dB?

What do you think about Jim's definition of the power ratio
being (feedline losses)/(feedline losses + load power)?
--
73, Cecil http://www.qsl.net/w5dxp

On Wed, 01 Dec 2004 20:33:14 -0600, Cecil Moore
wrote:
what is the feedline loss in dB?
The rest of the world computes it at "per 100 feet." Why change?

50 watts is entering the line, and 25 watts is exiting. The line loss is
3 dB. Surely you're able to make this calculation yourself.

I don't feel a need for a definition of a "power ratio". What you've
defined is indeed the ratio of two powers, but it escapes me of what use
it is except perhaps to cause confusion.

Roy Lewallen, W7EL

Cecil Moore wrote:

Roy Lewallen wrote:

Cecil Moore wrote:

Why is it so hard to get a straight answer to this simple question?


The answer is simple. 25 watts.


That was a given, Roy. The question is: what is the feedline
loss in dB?

What do you think about Jim's definition of the power ratio
being (feedline losses)/(feedline losses + load power)?
--
73, Cecil http://www.qsl.net/w5dxp

You really don't know?

The power into the input end of the transmission line is 50 watts.
(Surely you can subtract the circulator resistor power from the source
power to find the power entering the transmission line. Can't you?)

The power exiting the load end of the feedline is 25 watts.

Therefore the transmission line loss is 25 watts.

It does seem that you've gotten yourself confused by your bouncing waves
of average power.

Roy Lewallen, W7EL

Cecil Moore wrote:
Roy Lewallen wrote:

Line loss itself is a comparison of two powers, those entering and
leaving the line. The line loss in watts is the power entering the
line at the input end minus the power leaving the line at the output end.


Here's an example: The source is a signal generator equipped with a
circulator-resistor that dissipates all reflected power.

100w SGCR--------------feedline------------------mismatched load

The signal generator is sourcing 100 watts. The load is dissipating
25 watts. The circulator resistor is dissipating 50 watts. The
feedline is dissipating 25 watts. What is the feedline loss in dB?

Is that 25 watts lost from the signal generator output power of
100 watts or lost from the NET power available which is 50 watts?
--
73, Cecil http://www.qsl.net/w5dxp

Roy Lewallen wrote:

You really don't know?

You really don't know how to calculate the dB loss in the feedline?
If you do, why haven't you done so?

The power into the input end of the transmission line is 50 watts.
(Surely you can subtract the circulator resistor power from the source
power to find the power entering the transmission line. Can't you?)

That's *NET* power. The power into the transmission line is the 100w of
measured source power. The power dissipated in the circulator resistor is
the measured power out of the transmission line reflected from the mismatched
load. If you say to calculate the dB loss in the feedline based on the NET
power, that's what I will do.

I am trying to understand the difference in Bob's results and the
results using the ARRL equations. The magnitude of NET power to which
the feedline losses are ratio'ed may be the key to understanding that
difference.

The power exiting the load end of the feedline is 25 watts.

That was given.

Therefore the transmission line loss is 25 watts.

That was given but didn't answer the question about dB.

It does seem that you've gotten yourself confused by your bouncing waves
of average power.

Nope, you seem to be confused about what the question was. You keep
answering with what was given in the original example. The question
is: What is the *dB* loss in the feedline? Using NET power input, the
dB loss in the feedline is 3 dB, i.e. half of (100w-50w) = 50 watts.
I'm satisfied with that definition but an wondering why nobody else
has said the feedline losses equal 3 dB. What was so difficult about
that?

Incidentally, that figure agrees with Jim's equation which doesn't
even mention source power.
--
73, Cecil http://www.qsl.net/w5dxp